How to integrate K/[(y^2 + K)^3/2] dy

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In summary, the conversation discusses the integral of the function K/[(y2 + K)1.5] with respect to dy, which is used to find the electric field due to a line of uniformly charged wire. The textbook provides the integral as y/[d(y2 + k2)0.5], but there is confusion on how this answer was obtained. After some discussion, it is realized that the product rule was misused and the correct answer can be found by multiplying the integral by K.
  • #1
animboy
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As you can see the function given is

K/[(y2 + K)1.5]

The integral of this equation with respect to dy is part of the explanation my textbook gives of finding electric field due to line of uniformly charged wire, where y of course is the variable length of the wire. Then it gives the integral as

y/[d(y2 + k2)0.5]

but I can't see how it got this, since I try to differentiate this answer and got back this function which was different to the original.

Can someone whow me how the textbook got that answer? Thanks.
This function is radically different

Here is a picture of the original textbook statement.

[PLAIN]http://desmond.imageshack.us/Himg97/scaled.php?server=97&filename=physicsq.png&res=medium
 
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  • #2
integrate ((y^2 + K)^-3/2)dy=dv to get -((y^2 + K)^-1/2)/y=v

then

multiply that times K and you should be able to get the right answer since K is constant right?

this is just what I think
 
  • #3
Never mind, I found out what I did wrong. I misused the product rule. What a simple mistake.
 

1. What is the purpose of integrating K/[(y^2 + K)^3/2] dy?

The purpose of integrating K/[(y^2 + K)^3/2] dy is to find the antiderivative or the original function that, when differentiated, would result in the given equation. This process is also known as indefinite integration or antidifferentiation.

2. What are the steps to integrate K/[(y^2 + K)^3/2] dy?

The steps to integrate K/[(y^2 + K)^3/2] dy are as follows:

  1. Use the power rule to move the exponent in the denominator to the numerator and subtract 1 from the exponent. This will give you K(y^2 + K)^(-5/2).
  2. Use the chain rule to find the derivative of the expression inside the parentheses, which is 2y.
  3. Substitute the expression and its derivative into the integration formula for u-substitution, where u = y^2 + K and du = 2y dy.
  4. Integrate the new expression using the power rule, which will give you K/2(u)^(-5/2).
  5. Substitute back in the original expression for u and solve for the final answer.

3. Are there any special cases to consider when integrating K/[(y^2 + K)^3/2] dy?

Yes, when integrating K/[(y^2 + K)^3/2] dy, it is important to consider the case where K = 0. In this case, the original equation becomes simply 1/y^3, and the integration steps would change accordingly.

4. Can K/[(y^2 + K)^3/2] dy be solved using other integration techniques?

Yes, K/[(y^2 + K)^3/2] dy can also be solved using trigonometric substitutions or integration by parts. However, the method of u-substitution is the most commonly used and straightforward technique for this particular equation.

5. What are some real-world applications of integrating K/[(y^2 + K)^3/2] dy?

Integrating K/[(y^2 + K)^3/2] dy has various real-world applications in fields such as physics, engineering, and economics. For example, it can be used to calculate the work done by a varying force or to determine the velocity of an object undergoing acceleration. In economics, it can be used to model the relationship between supply and demand or to analyze the cost function of a business.

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